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Microscope and Microscopy - Where do we use Microscopes?-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Microscopy is the technical field of using microscopes to view objects and areas of objects that cannot be seen with the naked eye. In Grade 9 Biology, this is fundamental for studying the cell, the basic unit of life.

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Magnification (MM) is the process of enlarging the apparent size of an object. Total magnification in a compound microscope is the product of the magnification of the objective lens and the ocular lens: Mtotal=Mobjective×MocularM_{total} = M_{objective} \times M_{ocular}.

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Resolving Power (Resolution) is the ability of an optical instrument to show two close objects as separate. The limit of resolution (dd) is determined by the wavelength of light (λ\lambda) and the numerical aperture (NANA).

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The Compound Light Microscope is used in laboratories to observe living cells, stained sections of tissues, and microorganisms. It uses visible light (400 nm400 \text{ nm} to 700 nm700 \text{ nm}) and glass lenses.

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The Electron Microscope (EM) uses a beam of electrons instead of light. Because the wavelength of electrons is much shorter than that of visible light, EM can achieve magnifications up to 2,000,000×2,000,000 \times, allowing us to see organelles like ribosomes and DNA.

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Applications in Microbiology: Identifying pathogens like bacteria (1−10 μm1-10 \: \mu m) and viruses (20−300 nm20-300 \: nm). For example, observing the structure of Escherichia coliEscherichia \: coli requires high resolution.

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Applications in Histopathology: Surgeons and pathologists use microscopy to examine biopsy samples to detect cancerous cells, which often show abnormal nuclear-to-cytoplasmic ratios (N:CN:C ratio).

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Forensic Science: Using microscopes to analyze trace evidence such as hair, fibers, or glass fragments found at a crime scene.

📐Formulae

Mtotal=Mobjective×MeyepieceM_{total} = M_{objective} \times M_{eyepiece}

d=0.61×λNAd = \frac{0.61 \times \lambda}{NA}

NA=nsin⁡θNA = n \sin \theta

Actual Size=Image SizeMagnification\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}

💡Examples

Problem 1:

A student is using a compound microscope with a 10×10 \times eyepiece and a 45×45 \times objective lens. What is the total magnification of the specimen being observed?

Solution:

Given: Meyepiece=10×M_{eyepiece} = 10 \times Mobjective=45×M_{objective} = 45 \times

Using the formula: Mtotal=Mobjective×MeyepieceM_{total} = M_{objective} \times M_{eyepiece} Mtotal=45×10=450×M_{total} = 45 \times 10 = 450 \times

The total magnification is 450×450 \times.

Explanation:

The total magnification is calculated by multiplying the magnifying power of the two lenses the light passes through.

Problem 2:

An organelle is measured to be 0.5 cm0.5 \text{ cm} in a micrograph taken at a magnification of 50,000×50,000 \times. Calculate the actual size of the organelle in micrometers (μm\mu m).

Solution:

Given: Image Size=0.5 cm=5 mm=5000 μm\text{Image Size} = 0.5 \text{ cm} = 5 \text{ mm} = 5000 \: \mu m Magnification=50,000×\text{Magnification} = 50,000 \times

Using the formula: Actual Size=Image SizeMagnification\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}} Actual Size=5000 μm50,000\text{Actual Size} = \frac{5000 \: \mu m}{50,000} Actual Size=0.1 μm\text{Actual Size} = 0.1 \: \mu m

Converting to nanometers if needed: 0.1×1000=100 nm0.1 \times 1000 = 100 \text{ nm}.

Explanation:

To find the real size, we divide the measured size in the picture by the magnification factor, ensuring units are converted appropriately (1 cm=10,000 μm1 \text{ cm} = 10,000 \: \mu m).